Problem: Express $z_1=4\left[\cos\left(\dfrac{11\pi}{6}\right)+i\sin\left(\dfrac{11\pi}{6}\right)\right]$ in rectangular form. Express your answer in exact terms. $z_1=$
Explanation: The Strategy A complex number in rectangular form, $z={a}+{b}i$, can be written in polar form as $z={r}[\cos{\theta}+i\sin{\theta}]$, where ${r}$ is the absolute value, or modulus, and ${\theta}$ is the angle, or argument. Therefore, ${r}$ and ${\theta}$ can be found using the following formulas: ${r}=\sqrt{{a}^2+{b}^2}$ $\tan{\theta}=\dfrac{{b}}{{a}}$ [How did we get these equations?] Similarly, a complex number in polar form, $z={r}[\cos{\theta}+i\sin{\theta}]$, can be written in rectangular form as $z={a}+{b}i$, using the following formulas: ${a}={r}\cos{\theta}$ ${b}={r}\sin{\theta}$ [How did we get these equations?] Finding $a$ For $z_1={4}[\cos{\dfrac{11\pi}{6}}+i\sin{\dfrac{11\pi}{6}}]$ : ${r}={4}$ ${\theta}={\dfrac{11\pi}{6}}$ Therefore, we can find ${a}$ as follows. $\begin{aligned}{a}&={r}\cos{\theta} \\\\&={4}\cos{\dfrac{11\pi}{6}} \\\\&={2\sqrt{3}}\end{aligned}$ Finding $b$ $\begin{aligned}{b}&={r}\sin{\theta} \\\\&={4}\sin{\dfrac{11\pi}{6}} \\\\&={-2}\end{aligned}$ Summary $z_1={2\sqrt{3}}{-2}i$